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In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$. The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), so it admits a continuous right inverse (by Bartle-Graves Selection Theorem) but it may fail to have a bounded linear right inverse (an "extension operator"). Equivalently, $\text{ker}R_Y=\{f\in C(X): f_{|Y}=0\}$ may fail to be complemented in $C(X)$.

The possibly simpler counterexample is mentioned here, with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$.

Notably, here the closed set $Y$ is not metrisable. In fact (Luther and Martin, 1974) a metrisable $G_\delta$ subset $Y$ of $X$ does admit an extension operator --this generalises the classical Borsuk-Dugundji theorem, where $X$ is a metric space.

Metrisability of $Y$ is certainly not necessary, in general (e.g. in the quite trivial case of a clopen subset $Y$). Being a $G_\delta$ subset, that is, the zero set of a continuous function on $X$, is also not necessary, in general (e.g. in the quite trivial case of $Y=\{x_0\}$, where $x_0\in\ X$ does not have a countable basis of neighborhoods).

Question 1. Can we weaken the metrisability condition from Luther-Martin theorem, that is, what more general conditions on a closed set $Y$ ensure the existence of an extension operator?

The reason why in the above example with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$ the restriction operator $C(X)\to C(Y)$ is not a left-inverse is that $C(X)\sim\ell_\infty$ has a $w^*$-separable dual, whereas $C(Y)\sim \ell_\infty/c_0$ has not. This seems quite difficult to translate into more direct topological properties of $(X,Y)$. So it may be difficult to find a topological condition that guarantees the existence of an extension operator.

Question 2. Given a nbd $U$ of a closed set $Y$, can we find a closed set $Y'$ that admits an extension operator, $Y\subset Y'\subset U$?

Question 3. Suppose $Y_1$ and $Y_2$ admits an extension operator. Does $Y_1\cap Y_2$ admit an extension operator too?

Rmk. An extension operator $E:C(Y)\to C(X)$ is always of the form $Ef(x)= \langle Px,f\rangle$, where $P:X\to C(Y)^*$ is continuous w.r.to the $w^*$ topology of $C(Y)^*$ and $Py=\delta_y$ for all $y\in Y$.

In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$. The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), so it admits a continuous right inverse (by Bartle-Graves Selection Theorem) but it may fail to have a bounded linear right inverse (an "extension operator"). Equivalently, $\text{ker}R_Y=\{f\in C(X): f_{|Y}=0\}$ may fail to be complemented in $C(X)$.

The possibly simpler counterexample is mentioned here, with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$.

Notably, here the closed set $Y$ is not metrisable. In fact (Lutzer and Martin, 1974) a metrisable $G_\delta$ subset $Y$ of $X$ does admit an extension operator --this generalises the classical Borsuk-Dugundji theorem, where $X$ is a metric space.

Metrisability of $Y$ is certainly not necessary, in general (e.g. in the quite trivial case of a clopen subset $Y$). Being a $G_\delta$ subset, that is, the zero set of a continuous function on $X$, is also not necessary, in general (e.g. in the quite trivial case of $Y=\{x_0\}$, where $x_0\in\ X$ does not have a countable basis of neighborhoods).

Question 1. Can we weaken the metrisability condition from Lutzer-Martin theorem, that is, what more general conditions on a closed set $Y$ ensure the existence of an extension operator?

The reason why in the above example with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$ the restriction operator $C(X)\to C(Y)$ is not a left-inverse is that $C(X)\sim\ell_\infty$ has a $w^*$-separable dual, whereas $C(Y)\sim \ell_\infty/c_0$ has not. This seems quite difficult to translate into more direct topological properties of $(X,Y)$. So it may be difficult to find a topological condition that guarantees the existence of an extension operator.

Question 2. Given a nbd $U$ of a closed set $Y$, can we find a closed set $Y'$ that admits an extension operator, $Y\subset Y'\subset U$?

Question 3. Suppose $Y_1$ and $Y_2$ admits an extension operator. Does $Y_1\cap Y_2$ admit an extension operator too?

Rmk. An extension operator $E:C(Y)\to C(X)$ is always of the form $Ef(x)= \langle Px,f\rangle$, where $P:X\to C(Y)^*$ is continuous w.r.to the $w^*$ topology of $C(Y)^*$ and $Py=\delta_y$ for all $y\in Y$.

In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$. The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), but it may fail to have a bounded linear right inverse (an "extension operator" or “extensor”). Equivalently, $\text{ker}R_Y=\{f\in C(X): f_{|Y}=0\}$ may fail to be complemented in $C(X)$.

The possibly simpler counterexample is mentioned here, with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$.

Notably, here the closed set $Y$ is not metrisable. In fact (Luther and Martin, 1974) a metrisable $G_\delta$ subset $Y$ of $X$ does admit an extension operator --this generalises the classical Borsuk-Dugundji theorem, where $X$ is a metric space.

Metrisability of $Y$ is certainly not necessary, in general (e.g. in the quite trivial case of a clopen subset $Y$). Being a $G_\delta$ subset, that is, the zero set of a continuous function on $X$, is also not necessary, in general (e.g. in the quite trivial case of $Y=\{x_0\}$, where $x_0\in\ X$ does not have a countable basis of neighborhoods).

Question 1. Can we weaken the metrisability condition from Luther-Martin theorem, that is, what more general conditions on a closed set $Y$ ensure the existence of an extensor?

The reason why in the above example with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$ the restriction operator $C(X)\to C(Y)$ is not a left-inverse is that $C(X)\sim\ell_\infty$ has a $w^*$-separable dual, whereas $C(Y)\sim \ell_\infty/c_0$ has not. This seems quite difficult to translate into more direct topological properties of $(X,Y)$. So it may be difficult to find a topological condition that guarantees the existence of an extensor.

Question 2. Given a nbd $U$ of a closed set $Y$, can we find a closed set $Y'$ that admits an extensor, $Y\subset Y'\subset U$?

Rmk. An extensor $E:C(Y)\to C(X)$ is always of the form $Ef(x)= \langle Px,f\rangle$, where $P:X\to C(Y)^*$ is continuous w.r.to the $w^*$ topology of $C(Y)^*$ and $Py=\delta_y$ for all $y\in Y$.

In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$. The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), so it admits a continuous right inverse (by Bartle-Graves Selection Theorem) but it may fail to have a bounded linear right inverse (an "extension operator"). Equivalently, $\text{ker}R_Y=\{f\in C(X): f_{|Y}=0\}$ may fail to be complemented in $C(X)$.

The possibly simpler counterexample is mentioned here, with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$.

Notably, here the closed set $Y$ is not metrisable. In fact (Luther and Martin, 1974) a metrisable $G_\delta$ subset $Y$ of $X$ does admit an extension operator --this generalises the classical Borsuk-Dugundji theorem, where $X$ is a metric space.

Metrisability of $Y$ is certainly not necessary, in general (e.g. in the quite trivial case of a clopen subset $Y$). Being a $G_\delta$ subset, that is, the zero set of a continuous function on $X$, is also not necessary, in general (e.g. in the quite trivial case of $Y=\{x_0\}$, where $x_0\in\ X$ does not have a countable basis of neighborhoods).

Question 1. Can we weaken the metrisability condition from Luther-Martin theorem, that is, what more general conditions on a closed set $Y$ ensure the existence of an extension operator?

The reason why in the above example with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$ the restriction operator $C(X)\to C(Y)$ is not a left-inverse is that $C(X)\sim\ell_\infty$ has a $w^*$-separable dual, whereas $C(Y)\sim \ell_\infty/c_0$ has not. This seems quite difficult to translate into more direct topological properties of $(X,Y)$. So it may be difficult to find a topological condition that guarantees the existence of an extension operator.

Question 2. Given a nbd $U$ of a closed set $Y$, can we find a closed set $Y'$ that admits an extension operator, $Y\subset Y'\subset U$?

Question 3. Suppose $Y_1$ and $Y_2$ admits an extension operator. Does $Y_1\cap Y_2$ admit an extension operator too?

Rmk. An extension operator $E:C(Y)\to C(X)$ is always of the form $Ef(x)= \langle Px,f\rangle$, where $P:X\to C(Y)^*$ is continuous w.r.to the $w^*$ topology of $C(Y)^*$ and $Py=\delta_y$ for all $y\in Y$.

In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$. The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), but it may fail to have a bounded linear right inverse (an "extension operator"). Equivalently, $\text{ker}R_Y=\{f\in C(X): f_{|Y}=0\}$ may fail to be complemented in $C(X)$.

The possibly simpler counterexample is mentioned here, with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$.

Notably, here the closed set $Y$ is not metrisable. In fact (Luther and Martin, 1974) a metrisable $G_\delta$ subset $Y$ of $X$ does admit an extension operator --this generalises the classical Borsuk-Dugundji theorem, where $X$ is a metric space.

Metrisability of $Y$ is certainly not necessary, in general (e.g. in the quite trivial case of a clopen subset $Y$). Being a $G_\delta$ subset, that is, the zero set of a continuous function on $X$, is also not necessary, in general (e.g. in the quite trivial case of $Y=\{x_0\}$, where $x_0\in\ X$ does not have a countable basis of neighborhoods).

Question 1. Can we weaken the metrisability condition from Luther-Martin theorem, that is, what more general conditions on a closed set $Y$ ensure the existence of an extensor?

The reason why in the above example with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$ the restriction operator $C(X)\to C(Y)$ is not a left-inverse is that $C(X)\sim\ell_\infty$ has a $w^*$-separable dual, whereas $C(Y)\sim \ell_\infty/c_0$ has not. This seems quite difficult to translate into more direct topological properties of $(X,Y)$. So it may be difficult to find topological condition that guarantees the existence of an extensor.

Question 2. Given a nbd $U$ of a closed set $Y$, can we find a closed set $Y'$ that admits an extensor, $Y\subset Y'\subset U$?

Rmk. An extensor $E:C(Y)\to C(X)$ is always of the form $Ef(x)= \langle Px,f\rangle$, where $P:X\to C(Y)^*$ is continuous w.r.to the $w^*$ topology of $C(Y)^*$ and $Py=\delta_y$ for all $y\in Y$.

In the following $X$ is a Hausdorff compact topological space. Let $Y$ be a closed subset of $X$. The restriction operator $R_Y:C(X)\to C(Y)$ is surjective (Tietze), but it may fail to have a bounded linear right inverse (an "extension operator" or “extensor”). Equivalently, $\text{ker}R_Y=\{f\in C(X): f_{|Y}=0\}$ may fail to be complemented in $C(X)$.

The possibly simpler counterexample is mentioned here, with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$.

Notably, here the closed set $Y$ is not metrisable. In fact (Luther and Martin, 1974) a metrisable $G_\delta$ subset $Y$ of $X$ does admit an extension operator --this generalises the classical Borsuk-Dugundji theorem, where $X$ is a metric space.

Metrisability of $Y$ is certainly not necessary, in general (e.g. in the quite trivial case of a clopen subset $Y$). Being a $G_\delta$ subset, that is, the zero set of a continuous function on $X$, is also not necessary, in general (e.g. in the quite trivial case of $Y=\{x_0\}$, where $x_0\in\ X$ does not have a countable basis of neighborhoods).

Question 1. Can we weaken the metrisability condition from Luther-Martin theorem, that is, what more general conditions on a closed set $Y$ ensure the existence of an extensor?

The reason why in the above example with $X:=\beta\mathbb N$ and $Y:=\beta\mathbb N\setminus\mathbb N$ the restriction operator $C(X)\to C(Y)$ is not a left-inverse is that $C(X)\sim\ell_\infty$ has a $w^*$-separable dual, whereas $C(Y)\sim \ell_\infty/c_0$ has not. This seems quite difficult to translate into more direct topological properties of $(X,Y)$. So it may be difficult to find a topological condition that guarantees the existence of an extensor.

Question 2. Given a nbd $U$ of a closed set $Y$, can we find a closed set $Y'$ that admits an extensor, $Y\subset Y'\subset U$?

Rmk. An extensor $E:C(Y)\to C(X)$ is always of the form $Ef(x)= \langle Px,f\rangle$, where $P:X\to C(Y)^*$ is continuous w.r.to the $w^*$ topology of $C(Y)^*$ and $Py=\delta_y$ for all $y\in Y$.